Question

让我们处理这个数据样本

timeseries<-structure(list(Data = structure(c(10L, 14L, 18L, 22L, 26L, 29L, 
32L, 35L, 38L, 1L, 4L, 7L, 11L, 15L, 19L, 23L, 27L, 30L, 33L, 
36L, 39L, 2L, 5L, 8L, 12L, 16L, 20L, 24L, 28L, 31L, 34L, 37L, 
40L, 3L, 6L, 9L, 13L, 17L, 21L, 25L), .Label = c("01.01.2018", 
"01.01.2019", "01.01.2020", "01.02.2018", "01.02.2019", "01.02.2020", 
"01.03.2018", "01.03.2019", "01.03.2020", "01.04.2017", "01.04.2018", 
"01.04.2019", "01.04.2020", "01.05.2017", "01.05.2018", "01.05.2019", 
"01.05.2020", "01.06.2017", "01.06.2018", "01.06.2019", "01.06.2020", 
"01.07.2017", "01.07.2018", "01.07.2019", "01.07.2020", "01.08.2017", 
"01.08.2018", "01.08.2019", "01.09.2017", "01.09.2018", "01.09.2019", 
"01.10.2017", "01.10.2018", "01.10.2019", "01.11.2017", "01.11.2018", 
"01.11.2019", "01.12.2017", "01.12.2018", "01.12.2019"), class = "factor"), 
    client = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L), .Label = c("Horns", "Kornev"), class = "factor"), stuff = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 
    3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("chickens", 
    "hooves", "Oysters"), class = "factor"), Sales = c(374L, 
    12L, 120L, 242L, 227L, 268L, 280L, 419L, 12L, 172L, 336L, 
    117L, 108L, 150L, 90L, 117L, 116L, 146L, 120L, 211L, 213L, 
    67L, 146L, 118L, 152L, 122L, 201L, 497L, 522L, 65L, 268L, 
    441L, 247L, 348L, 445L, 477L, 62L, 226L, 476L, 306L)), .Names = c("Data", 
"client", "stuff", "Sales"), class = "data.frame", row.names = c(NA, 
-40L))

我想按组使用auto.arima进行预测

# first the grouping variable
timeseries$group <- paste0(timeseries$client,timeseries$stuff)

# now the list
listed <- split(timeseries,timeseries$group)

library("forecast")
library("lubridate")

listed_ts <- lapply(listed,
                    function(x) ts(x[["Sales"]], start = start = c(2017, 1), frequency = 12)  ) 

listed_ts

listed_arima <- lapply(listed_ts,function(x) auto.arima(x) )
#Now the forecast for each arima:
listed_forecast <- lapply(listed_arima,function(x) forecast(x,2) )
listed_forecast
do.call(rbind,listed_forecast)

如果这样做的话，我会对未来进行预测，但是我想看看，auto.arima模型根据我的示例预测的初始值是什么。更加清楚。在我的示例中，Sales代表01.04.2017角鸡= 374。对？如何从示例数据中看到该日期和另一个日期的auto.arima模型预测的值。

所以输出

Answer 1

这些值称为 fitted 值，可以使用函数fitted如下获得它们：

lapply(listed_arima, fitted)
# $Hornschickens
#          Jan      Feb      Mar      Apr      May      Jun      Jul      Aug      Sep      Oct      Nov
# 2017 223.8182 223.8182 223.8182 223.8182 223.8182 223.8182 223.8182 223.8182 223.8182 223.8182 223.8182
#
# $Hornshooves
#           Jan      Feb      Mar      Apr      May      Jun      Jul      Aug      Sep      Oct      Nov      Dec
# 2017 336.9231 336.9231 336.9231 336.9231 336.9231 336.9231 336.9231 336.9231 336.9231 336.9231 336.9231 336.9231
# 2018 336.9231                                                                                                   
#
# $KornevOysters
#          Jan     Feb     Mar     Apr     May     Jun     Jul     Aug     Sep     Oct     Nov     Dec
# 2017 137.125 137.125 137.125 137.125 137.125 137.125 137.125 137.125 137.125 137.125 137.125 137.125
# 2018 137.125 137.125 137.125 137.125

在这种情况下，结果并不十分有趣，因为所有拟合模型都是ARIMA（0,0,0）-白噪声。

请注意，该解决方案等效于

lapply(listed_arima, function(x) fitted(x))

出于相同的原因，您也可以使用

listed_arima <- lapply(listed_ts, auto.arima)

在R中使用auto.arima显示初始数据的预测值

1 个答案: